Question

Consider N = n₁n₂ identical cells, each of emf ε and internal resistance r. Suppose n₁ cells are joined in series to form a line and n₂ such lines are connected in parallel.

The combination drives a current in an external resistance R. (a) Find the current in the external resistance. (b) Assuming that n₁ and n₂ can be continuously varied, find the relation between n₁, n₂, R and r for which the current in R is maximum.

Answer 1

(a)

Given:-

Emf of one cell = E

∴ Total e.m.f. of n₁ cells in one row = n₁E

Total emf of one row will be equal to the net emf across all the n₂ rows because of parallel connection.

Total resistance in one row = n₁r

Total resistance of n₂ rows in parallel \[= \frac{n_1 r}{n_2}\]

Net resistance of the circuit = R + \[\frac{n_1 r}{n_2}\]

\[\therefore \text{Current, }  I =   \frac{n_1 E}{R + \frac{n_1 r}{n_2}} = \frac{n_1 n_2 E}{n_2 R + n_1 r}\]

 

(b) From (a),

\[I = \frac{n_1 n_2 E}{n_2 R + n_1 r}\]

For I to be maximum, (n₁r + n₂R) should be minimum

\[\Rightarrow  \left( \sqrt{n_1 r} - \sqrt{n_2 R} \right)^2  + 2\sqrt{n_1 R  n_2 r} = \min\]

It is minimum when

\[\sqrt{n_1 r} = \sqrt{n_2 R}\]

\[       n_1 r =  n_2 R\]

∴ I is maximum when n₁r = n₂R .